THEORYMECHANICS902Chapter13 Theorem ofkinetic energyCollege of Mechanical and VehicleEngineering王晓君
College of Mechanical and Vehicle Engineering 王晓君 Chapter13 Theorem of kinetic energy THEORY MECHANICS
Work of forceThe kinetic energy of the particleand its systemTheorem of kinetic energyExamples of comprehensiveapplication of universal theorem
• Work of force • The kinetic energy of the particle and its system • Theorem of kinetic energy • Examples of comprehensive application of universal theorem
LntroductionThe first two chapters established the relationship between thephysical activity change of particle and particles system and theexternal force and acting time.Based on work and kinetic energy, this chapter establishes therelationship between the change of kinetic energy of a particle orsystem of particles and the work of the force, namely kinetic energytheorem.Differentfrom the momentum theorem and the moment ofmomentum theorem, the kinetic energy theorem is more convenientand effective to analyze the dynamics of particles and particlesystems from the perspective of energy. It also can establish theconnection between the mechanical movement and other forms ofmovement.Before the introduction to the kinetic energy theorem, firstintroduced the relevant physical quantities: work with kinetic energy
Introduction The first two chapters established the relationship between the physical activity change of particle and particles system and the external force and acting time. Based on work and kinetic energy, this chapter establishes the relationship between the change of kinetic energy of a particle or system of particles and the work of the force, namely kinetic energy theorem. Different from the momentum theorem and the moment of momentum theorem, the kinetic energy theorem is more convenient and effective to analyze the dynamics of particles and particle systems from the perspective of energy. It also can establish the connection between the mechanical movement and other forms of movement. Before the introduction to the kinetic energy theorem, first introduced the relevant physical quantities: work with kinetic energy
I .The concept of work1,The work of the normal force13.1Suppose the object travels along a straight方Work of forcepath S under the action of constant forceFLM' M.α αas shown in the figure, then the work doneby the force Wis defined asW=Fcosα.s= F.3Work is algebraic quantity. It represents the cumulativeeffect of a force over a distance, so work is the cumulativeeffect.In the SI system of units, the unit of work is J(Joule). 1J = 1N . mM2A2, The work of the variable forcedsaijaMLet the particle M move along the curveunder the action of variable force F, as shown万in the figure, and the work done by the force F[M,on the tiny arc is called the element work ofthe force, denoted as SwW
Ⅰ.The concept of work M1 M2 s F F 1、The work of the normal force Suppose the object travels along a straight path S under the action of constant force , as shown in the figure, then the work done by the force W is defined as F W F s F s = cos = Work is algebraic quantity. It represents the cumulative effect of a force over a distance, so work is the cumulative effect. In the SI system of units, the unit of work is J (Joule). 1J =1N m 2、The work of the variable force M1 M M2 M F dsdr Let the particle M move along the curve under the action of variable force , as shown in the figure, and the work done by the force on the tiny arc is called the element work of the force, denoted as F F W 13.1 Work of force
I .The concept of workLet the displacement of the correspondingM,small arc ds be dr then the expression ofdsVaaMelement work of the force is W = F.drFW, M .dr--- adus ctor ethodMJM,SW = Fcosα·ds = F:dsM?Mn? Fcosα·ds :Wi2 =F.ds ---- The natural methodJM,JM,dr = dxi + dyi + dzkSuppose: F = Xi +Yi+ ZkSW = Xdx + Ydy + ZdzMWi2 =Xdx + Ydy + ZdzJMRectangularcoordinate method
W F ds F ds M M M M = = 2 1 2 1 cos 1 2 M1 M M2 M F dsdr Suppose： F Xi Yj Zk = + + dr dxi dyj dzk = + + W = Xdx +Ydy + Zdz = + + 2 1 1 2 M M W Xdx Ydy Zdz W = F ds = F ds cos = 2 1 12 M M W F dr Let the displacement of the corresponding small arc be , then the expression of element work of the force is ds dr W F dr = - The natural method - Radius vector method - Rectangular coordinate method Ⅰ.The concept of work
I .The concept of work3, The work of the resultant forceIf there are n forces acting on a particle, the netforce is R- ZE , When the particle moves from M, to M,the work done by the resultant force R isMM2 R.dr(m (E +F, +..+F).drWi2 = -JM,MMM2M2 F. .drA?dr + M? F, dr +..JM,JMUM=W +W, +...+W, =ZWSo, the work done by the resultant force in the convergentforce system at any distance is egual to the algebraic sumof the work done by the components at the same distance
3、The work of the resultant force If there are forces acting on a particle, the net force is , When the particle moves from to , the work done by the resultant force is M1 M2 R n i M M n M M M M n M M M M W W W W F dr F dr F dr W R dr F F F dr = + + + = = + + + = = + ++ 1 2 1 2 1 2 1 2 2 1 2 1 2 1 2 1 2 1 ( ) So, the work done by the resultant force in the convergent force system at any distance is equal to the algebraic sum of the work done by the components at the same distance. Ⅰ.The concept of work n R Fi =
II. The work of a common force1, The work of gravityLet the mass of the particle bem, it'sgoing to move from M, to M, under theZMVforce of gravity. Establish the coordinateM2Zas shown in the figure, X = O,Y = O,Z = -mgy22mgOByWi2= fmXdx+ Ydy +ZdzJMxWe have Wi2 = (2 (-mg)dz = mg(z1 - z2)For a system of particles, the work done by its gravity isW12 = Zm,g(zi1 - zi2) =(Zm,zi1 -Zm,zi2)g= (Mzcl - Mzc2)g = Mg(zc1 - zc2)Thus, the work of gravity depends only on the startingand ending position of the center of gravity, and hasnothing to do with the path the center of gravity travels
Ⅱ. The work of a common force 1、The work of gravity x y z M M2 M1 O 1 z 2 z mg Let the mass of the particle be ，it's going to move from to under the force of gravity. Establish the coordinate as shown in the figure, m M1 M2 X = 0,Y = 0,Z = −mg By = + + 2 1 1 2 M M W Xdx Ydy Zdz We have ( ) ( ) 1 2 1 2 2 1 W m g dz m g z z z z = − = − For a system of particles, the work done by its gravity is ( ) ( ) ( ) ( ) 1 2 1 2 12 1 2 1 2 C C C C i i i i i i i Mz Mz g Mg z z W m g z z m z m z g = − = − = − = − Thus, the work of gravity depends only on the starting and ending position of the center of gravity, and has nothing to do with the path the center of gravity travels
I. The work of a common force2, The work of the elastic forceF&Let the particle M move alongMMfrthe trajectory under the action ofKS0t2elastic forces, the original length ofMloOthe spring is lo, elasticity coefficientis k, in the elastic range, the elasticforce Fis F=-k(r-l.)rM2From Wiz = JM F.dr we have Wi2 = JM -k(r-lo)-drMMr2Wi2 = ["2 - k(r -lo)dr2=k[(ri -l。)2-(rz -l。)2]ork(82 -82)Wi2 =22
2、The work of the elastic force O M1 M2 M F 0 l 1 r 2 r r 0 r 1 2 Let the particle M move along the trajectory under the action of elastic forces, the original length of the spring is , elasticity coefficient is , in the elastic range, the elastic force is 0 l k F 0 0 F k(r l )r = − − From = 2 1 12 M M W F dr we have = − − 2 1 1 2 0 0 ( ) M M W k r l r dr 2 2 0 2 1 0 2 1 2 0 0 ( ) ( ) 2 1 ( ) 2 1 ( ) 2 1 2 1 k r l r l W k r l dr k r l r r r r = − − − = − − = − − or ( ) 2 1 2 2 2 W1 2 = k 1 − Ⅱ. The work of a common force
II. The work of a common force3, The work applied on a rigid body rotating in a fixed axisZThe force Facting on a fixed axis rotating rigidbody is shown in the figure. The work done by the百rigid body as it moves from positionPrto position PLis Wi2 =|M F,ds = [" F,-O,MdpJM00M.(F)= F ·OMOf whichMSoWi2 = [" M.(F)dp0When M.(F)is constant, W2 = M,(F)·(P2 -P,)If applied to a rigid body it is a couple of forces, torque vectoris m, then the work done by the couple isWi2 = [" m.dpWhen m, is constant,W2 = m,(P2 -P)
3、The work applied on a rigid body rotating in a fixed axis z O1 M O F F Fb Fn The force acting on a fixed axis rotating rigid body is shown in the figure. The work done by the rigid body as it moves from position to position is F 1 2 W F ds F O Md M M 12 1 2 1 2 1 = = So W M z (F)d 2 1 12 = When is constant, M (F) z ( ) ( ) 12 2 −1 W = M F z Of which Mz (F) = F O1 M If applied to a rigid body it is a couple of forces, torque vector is m , then the work done by the couple is W mz d = 2 1 12 When is constant, mz ( ) W12 = mz 2 −1 Ⅱ. The work of a common force
I. The work of a common forceZ4, The work of frictionmgAs shown in the figure, when the objectslides on a fixed surface, the work of slidingNXfriction is Wi2 = [-Fds =-{[ fNdsSWhenfN is constant, Wz =-fNsWhere s is the arc length of the particle. It follows that the workdone by friction depends on the path traveled by the particle.As shown in the figure, when the rigid body rollspurely on a fixed plane, the frictionalforce and theFnormalreaction force act on the instantaneous777C*/77777center. Since the velocity of the instantaneous centerNis equal to zero, there is no displacement in theinstantaneous center, so the frictionalforce and thenormalreaction forcedono work
4、The work of friction x y z O mg N F s M1 M2 As shown in the figure, when the object slides on a fixed surface, the work of sliding friction is = − = − 2 1 2 1 12 s s s s W Fds f Nds When is constant, f N W = − f Ns 12 Where is the arc length of the particle. It follows that the work done by friction depends on the path traveled by the particle. s N F C As shown in the figure, when the rigid body rolls purely on a fixed plane, the frictional force and the normal reaction force act on the instantaneous center. Since the velocity of the instantaneous center is equal to zero, there is no displacement in the instantaneous center, so the frictional force and the normal reaction force do no work. Ⅱ. The work of a common force